目录
一、数据炸裂
0 问题描述
如何将字符串1-5,16,11-13,9" 扩展成 "1,2,3,4,5,16,11,12,13,9" 且顺序不变。
1 数据准备
with data as (select '1-5,16,11-13,9' as a)2 数据分析
步骤一：explode(split(a, ',')) 炸裂 + row_number()排序，一行变多行，且对每行的数据排序，保证有序性。
with data as (select '1-5,16,11-13,9' as a)
select
a,
row_number() over () as rn
from (
select
explode(split(a, ',')) as a
from data
)tmp1;输出结果：
步骤二： lateral view explode(split(a, '-')) 、max(b) - min(b) as diff
（1）lateral view +explode 侧写和炸裂，一行变多行，并将源表中每行的输出结果与该行连接；
（2）group by a, rn ....... select min(b) as start_index 得到每个分组的起始值
（3）max(b) - min(b) 得到每个分组的步长
with data as (select '1-5,16,11-13,9' as a)
select
a,
rn,
min(b) as start_data,
max(b) - min(b) as diff
from (
select
a,
rn,
b
from (
select
a,
row_number() over () as rn
from (
select
explode(split(a, ',')) as a
from data
) tmp1
) tmp2
lateral view explode(split(a, '-')) table1 as b
) tmp3
group by a, rn;输出结果是：
步骤三： 根据步长生成索引值，起始值加上索引值获取展开值
（1） lateral view posexplode(split(space(cast (diff as int)), '')) table1 as pos, item;
侧写和炸裂，根据分组的步长 diff 生成对应的索引值pos
（2）(start_data + pos) as str,起始值加上索引值获取展开值
with data as (select '1-5,16,11-13,9' as a)
select
a,
rn,
cast ((start_data + pos) as int) as str
from (
select
a,
rn,
start_index,
diff,
pos
from (
select
a,
rn,
min(b) as start_data,
max(b) - min(b) as diff
from (
select
a,
rn,
b
from (
select
a,
row_number() over () as rn
from (
select
explode(split(a, ',')) as a
from data
) tmp1
) tmp2
lateral view explode(split(a, '-')) table1 as b
) tmp3
group by a, rn
) tmp4
lateral view posexplode(split(space(cast(diff as int)), '')) table1 as pos, val) tmp5
order by rn;输出结果是：
步骤四： 对a,rn, diff 字段分组，拼接str字符串得到最终结果值
with data as (select '1-5,16,11-13,9' as a)
select
concat_ws(',', collect_set(cast(str as string))) as result
from (
select
a,
rn,
cast((start_index + pos) as int) as str
from (
select
a,
rn,
start_index,
diff,
pos
from (
select
a,
rn,
min(b) as start_index,
max(b) - min(b) as diff
from (
select
a,
rn,
b
from (
select
a,
row_number() over () as rn
from (
select
explode(split(a, ',')) as a
from data
) tmp1
) tmp2
lateral view explode(split(a, '-')) table1 as b
) tmp3
group by a, rn
) tmp4
lateral view posexplode(split(space(cast(diff as int)), '')) table1 as pos, val
) tmp5
) tmp6
group by a,rn,diff;最终的输出结果：1,2,3,4,5,16,11,12,13,9
3 小结
数据炸裂的思路一般是：
1.计算区间【a,b】的步长(差值)diff；
2.利用split分割函数+ posexplode等 将一行变成 diff+1 行，生成对应的下角标pos(pos的取值为【0,diff】)；
3.【a,b】区间的起始值 （a + pos） 将数据平铺开；
4.基于平铺开后的数据集进一步加工处理，例如：分组聚合等。二、数据合并
0 问题描述
面试题：基于A表的数据生成B表数据
1 数据准备
create table if not exists tableA
(
id string comment '用户id',
name string comment '用户姓名'
) comment 'A表';
insert overwrite table tableA values
('1','aa'),
('2','aa'),
('3','aa'),
('4','d'),
('5','c'),
('6','aa'),
('7','aa'),
('8','e'),
('9','f'),
('10','g');
create table if not exists tableC
(
id string comment '用户id',
name string comment '用户姓名'
) comment 'C表';
insert overwrite table tableC values
('3','aa|aa|aa'),
('4','d'),
('5','c'),
('7','aa|aa'),
('8','e'),
('9','f'),
('10','g');
2 数据分析
步骤1：寻找满足条件的断点
select
id,
name,
if(name != lag_name, 1, 0) as flag
from (
select
id,
name,
lag(name, 1, name) over (order by cast(id as int)) as lag_name
from tableA
) tmp1;输出结果为:
步骤2：断点处标记为1，非断点处标记为0，并对断点标记值进行累加，构造分组标签
select
id,
name,
--并对断点标记值进行累加，构造分组标签
sum(flag) over (order by cast(id as int)) grp
from (
select
id,
name,
--断点处标记为1，非断点处标记为0
if(name != lag_name, 1, 0) flag
from (
select
id,
name,
lag(name, 1, name) over (order by cast(id as int)) as lag_name
from tableA
) tmp1
) tmp2;输出结果为：
步骤3：按照分组标签进行数据合并，并取得分组中最大值作为id
select
max_id,
-- collect_list 数据聚合并拼接concat_ws
concat_ws('|', collect_list(name)) as name
from (
select
name,
grp,
max(id) over (partition by grp) max_id
from (
select
id,
name,
sum(if(name != lag_name, 1, 0)) over (order by cast(id as int)) as grp
from (
select
id,
name,
lag(name, 1, name) over (order by cast(id as int)) as lag_name
from tableA
) tmp1
) tmp2
) tmp3
group by max_id, grp;输出结果为：
通过max_id, grp分组，对name进行 concat_ws('|', collect_list(name)) 聚合拼接，得出最终的结果
3 小结
断点分组问题的算法总结
步骤1：寻找满足条件的断点
步骤2：断点处标记值为1，非断点处标记为0
步骤3：对断点标记值进行累加 sum(xx)over(order by xx)，构造分组标签
步骤4：按照分组标签进行分组求解问题
